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THE POLYGON OF FORCES 39 
the beam AB, viz. U, T, and §, and since these forces are in equilibrium 
and are not parallel, their lines of action must meet at a point which 
must be the point N where the lines of action of U and T intersect. 
This determines the line of action of S, and the polygon of forces being 
completed, the magnitudes of T and § are found to be, T = 312°6 lbs., and 
Fig. 35. 
‘ 
S = 593°3 lbs., and 6, the angle which the line of action of S makes with 
the beam, is 26° 3’. 
The forces T and S may, however, be found without the use of the 
point N, as follows. Draw as much of the polygon of forces as the data 
of the problem will permit. Choose a pole o and draw the funicular 
polygon B5678, starting at the point B, which is the only point in the 
line of action of S which is as yet known. 8B is‘the closing side of this 
funicular polygon, and a line oc drawn parallel to 8B to meet that side 
of the polygon of forces which is parallel to the line of action of 'T will 
determine the remaining angular point of the polygon of forces, and will 
ere fix the magnitude of T and also the direction and magnitude 
of 8, . . 
58. Analytical Methods.—The following examples illustrate the 
methods of solving, by calculation, problems on forces acting in a plane 
and at a point. 
(1) Two forces, P = 20 Ibs., and Q= 10 lbs., act as shown at (a), Fig. 
36, the angle between their lines of action 
being 100°. It is required to find R, the FP so) P 
‘resultant of P and Q. > © ~ RQ 
Drawing the triangle of forces shown "Q (@) “(b) ~~ 
at ()), the angle opposite to R is Fic. 36. 
180° — 100° = 80°. 
Then, R? = P? + Q? — 2PQ cos 80° 
= 20? + 10? — 2 x 20 x 10 k 0°17365 = 430°54. 
Therefore, R= ,/430°54 = 20°75 Ibs. 
The angle @ which R makes with P is found from the equation 
sn@ Q 
in 80° R 
